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Consider the following statement: $\forall m \in \mathbb{N}: (m>1 \wedge m|n) \Rightarrow m=n$. What kind of element is n? This is my reasoning, is it ok?

$ (m>1 \wedge m|n) \Rightarrow m=n \Leftrightarrow m \le 1 \vee m \not | n \vee m=n \Leftrightarrow False \vee m \not | n \vee m=n \Leftrightarrow m \not | n \vee m = n$. Now, suppose $m \neq n \Rightarrow m \not | n$. So $n$ is not divisible by every natural number, except itself. So $n$ is a prime number.

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I'm confused about what set you're talking about. The set $\{ m \in \mathbb{N} | m>1, m|n \}$ is the set of divisors of $n$ greater than $1$. –  Dimitrije Kostic Dec 12 '11 at 17:56
    
Sorry for the confusion, doesn't mean the set, I mean the elements $n$ which apply to the statement above. –  Kevin Dec 12 '11 at 17:59
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up vote 2 down vote accepted

The reasoning is almost OK. It would have been better to express the reasoning in a less symbol-dense way. The identification of $m\le 1$ with False is wrong, since there is no reason to exclude $m=1$.

The sentence says that any divisor of $n$ greater than $1$ is equal to $n$. For $n>1$, this is equivalent to saying that $n$ is prime.

But note that $1$ has the property in question, since it is indeed true that every divisor of $1$ which is greater than $1$ is equal to $1$. (Every such divisor $m$ is also equal to $47$, since there are no such divisors.) So the elements $n$ of $\mathbb{N}$ that satisfy our formula are the primes together with $1$.

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Is $1$ in that set? It has no divisors greater than $1$. –  Dimitrije Kostic Dec 12 '11 at 18:22
    
@Dimitrije Kostic: Because $1$ has no divisors greater than $1$, $(m>1) \land (m|1)$ is false for every $m$. When $A$ is false, $A\implies B$ is true for any $B$. –  André Nicolas Dec 12 '11 at 18:32
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